Convert to Interval Notation |2x-6|+|3-x|>12

$| 2 x - 6 | + | 3 - x | > 12$

Replace $>$ with $=$ in $| 2 x - 6 | + | 3 - x | > 12$ .

$| 2 x - 6 | + | 3 - x | = 12$

Subtract $| 2 x - 6 |$ from both sides of the equation.

$| 3 - x | = 12 - | 2 x - 6 |$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$3 - x = \pm (12 - | 2 x - 6 |)$

The result consists of both the positive and negative portions of the $\pm$ .

$3 - x = 12 - | 2 x - 6 |$

$3 - x = - (12 - | 2 x - 6 |)$

Solve $3 - x = 12 - | 2 x - 6 |$ for $x$ .

Solve for $| 2 x - 6 |$ .

Rewrite the equation as $12 - | 2 x - 6 | = 3 - x$ .

$12 - | 2 x - 6 | = 3 - x$

Move all terms not containing $| 2 x - 6 |$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$- | 2 x - 6 | = 3 - x - 12$

Subtract $12$ from $3$ .

$- | 2 x - 6 | = - x - 9$

Divide each term in $- | 2 x - 6 | = - x - 9$ by $- 1$ and simplify.

Divide each term in $- | 2 x - 6 | = - x - 9$ by $- 1$ .

$\frac{- | 2 x - 6 |}{- 1} = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| 2 x - 6 |}{1} = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Divide $| 2 x - 6 |$ by $1$ .

$| 2 x - 6 | = \frac{- x}{- 1} + \frac{- 9}{- 1}$

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

$| 2 x - 6 | = \frac{x}{1} + \frac{- 9}{- 1}$

Divide $x$ by $1$ .

$| 2 x - 6 | = x + \frac{- 9}{- 1}$

Divide $- 9$ by $- 1$ .

$| 2 x - 6 | = x + 9$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 x - 6 = \pm (x + 9)$

The result consists of both the positive and negative portions of the $\pm$ .

$2 x - 6 = x + 9$

$2 x - 6 = - (x + 9)$

Solve $2 x - 6 = x + 9$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$2 x - 6 - x = 9$

Subtract $x$ from $2 x$ .

$x - 6 = 9$

Move all terms not containing $x$ to the right side of the equation.

Add $6$ to both sides of the equation.

$x = 9 + 6$

Add $9$ and $6$ .

$x = 15$

Solve $2 x - 6 = - (x + 9)$ for $x$ .

Simplify $- (x + 9)$ .

Rewrite.

$2 x - 6 = 0 + 0 - (x + 9)$

Simplify by adding zeros.

$2 x - 6 = - (x + 9)$

Apply the distributive property.

$2 x - 6 = - x - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$2 x - 6 = - x - 9$

Move all terms containing $x$ to the left side of the equation.

Add $x$ to both sides of the equation.

$2 x - 6 + x = - 9$

Add $2 x$ and $x$ .

$3 x - 6 = - 9$

Move all terms not containing $x$ to the right side of the equation.

Add $6$ to both sides of the equation.

$3 x = - 9 + 6$

Add $- 9$ and $6$ .

$3 x = - 3$

Divide each term in $3 x = - 3$ by $3$ and simplify.

Divide each term in $3 x = - 3$ by $3$ .

$\frac{3 x}{3} = \frac{- 3}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 3}{3}$

Divide $x$ by $1$ .

$x = \frac{- 3}{3}$

Simplify the right side.

Divide $- 3$ by $3$ .

$x = - 1$

Consolidate the solutions.

$x = 15, - 1$

Solve $3 - x = - (12 - | 2 x - 6 |)$ for $x$ .

Solve for $| 2 x - 6 |$ .

Rewrite the equation as $- (12 - | 2 x - 6 |) = 3 - x$ .

$- (12 - | 2 x - 6 |) = 3 - x$

Simplify $- (12 - | 2 x - 6 |)$ .

Apply the distributive property.

$- 1 \cdot 12 - - | 2 x - 6 | = 3 - x$

Multiply $- 1$ by $12$ .

$- 12 - - | 2 x - 6 | = 3 - x$

Multiply $- - | 2 x - 6 |$ .

Multiply $- 1$ by $- 1$ .

$- 12 + 1 | 2 x - 6 | = 3 - x$

Multiply $| 2 x - 6 |$ by $1$ .

$- 12 + | 2 x - 6 | = 3 - x$

Move all terms not containing $| 2 x - 6 |$ to the right side of the equation.

Add $12$ to both sides of the equation.

$| 2 x - 6 | = 3 - x + 12$

Add $3$ and $12$ .

$| 2 x - 6 | = - x + 15$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 x - 6 = \pm (- x + 15)$

The result consists of both the positive and negative portions of the $\pm$ .

$2 x - 6 = - x + 15$

$2 x - 6 = - (- x + 15)$

Solve $2 x - 6 = - x + 15$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Add $x$ to both sides of the equation.

$2 x - 6 + x = 15$

Add $2 x$ and $x$ .

$3 x - 6 = 15$

Move all terms not containing $x$ to the right side of the equation.

Add $6$ to both sides of the equation.

$3 x = 15 + 6$

Add $15$ and $6$ .

$3 x = 21$

Divide each term in $3 x = 21$ by $3$ and simplify.

Divide each term in $3 x = 21$ by $3$ .

$\frac{3 x}{3} = \frac{21}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{21}{3}$

Divide $x$ by $1$ .

$x = \frac{21}{3}$

Simplify the right side.

Divide $21$ by $3$ .

$x = 7$

Solve $2 x - 6 = - (- x + 15)$ for $x$ .

Simplify $- (- x + 15)$ .

Rewrite.

$2 x - 6 = 0 + 0 - (- x + 15)$

Simplify by adding zeros.

$2 x - 6 = - (- x + 15)$

Apply the distributive property.

$2 x - 6 = - - x - 1 \cdot 15$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$2 x - 6 = 1 x - 1 \cdot 15$

Multiply $x$ by $1$ .

$2 x - 6 = x - 1 \cdot 15$

Multiply $- 1$ by $15$ .

$2 x - 6 = x - 15$

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$2 x - 6 - x = - 15$

Subtract $x$ from $2 x$ .

$x - 6 = - 15$

Move all terms not containing $x$ to the right side of the equation.

Add $6$ to both sides of the equation.

$x = - 15 + 6$

Add $- 15$ and $6$ .

$x = - 9$

Consolidate the solutions.

$x = 7, - 9$

Consolidate the solutions.

$x = 15, - 1, 7, - 9$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 1$

$- 1 < x < 7$

$7 < x < 15$

$x > 15$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$| 2 (- 12) - 6 | + | 3 - (- 12) | > 12$

The left side $45$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $- 9 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < - 1$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$| 2 (- 5) - 6 | + | 3 - (- 5) | > 12$

The left side $24$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$| 2 (0) - 6 | + | 3 - (0) | > 12$

The left side $9$ is not greater than the right side $12$ , which means that the given statement is false.

False

Test a value on the interval $7 < x < 15$ to see if it makes the inequality true.

Choose a value on the interval $7 < x < 15$ and see if this value makes the original inequality true.

$x = 11$

Replace $x$ with $11$ in the original inequality.

$| 2 (11) - 6 | + | 3 - (11) | > 12$

The left side $24$ is greater than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $x > 15$ to see if it makes the inequality true.

Choose a value on the interval $x > 15$ and see if this value makes the original inequality true.

$x = 18$

Replace $x$ with $18$ in the original inequality.

$| 2 (18) - 6 | + | 3 - (18) | > 12$

The left side $45$ is greater than the right side $12$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 9$ True

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$7 < x < 15$ True

$x > 15$ True

$x < - 9$ True

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$7 < x < 15$ True

$x > 15$ True

The solution consists of all of the true intervals.

$x < - 9$ or $- 9 < x < - 1$ or $7 < x < 15$ or $x > 15$

Convert the inequality to interval notation.

$(- \infty, - 9) \cup (- 9, - 1) \cup (7, 15) \cup (15, \infty)$

Convert to Interval Notation |2x-6|+|3-x|>12

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